Question: Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}6x+8y &= -3 \\ -6x+5y &= 3\end{align*}$
Begin by moving the $x$ -term in the second equation to the right side of the equation. $5y = 6x+3$ Divide both sides by $5$ to isolate $y$ $y = {\dfrac{6}{5}x + \dfrac{3}{5}}$ Substitute this expression for $y$ in the first equation. $6x+8({\dfrac{6}{5}x + \dfrac{3}{5}}) = -3$ $6x + \dfrac{48}{5}x + \dfrac{24}{5} = -3$ Simplify by combining terms, then solve for $x$ $\dfrac{78}{5}x + \dfrac{24}{5} = -3$ $\dfrac{78}{5}x = -\dfrac{39}{5}$ $x = -\dfrac{1}{2}$ Substitute $-\dfrac{1}{2}$ for $x$ back into the top equation. $6( -\dfrac{1}{2})+8y = -3$ $-3+8y = -3$ $8y = 0$ $y = 0$ The solution is $\enspace x = -\dfrac{1}{2}, \enspace y = 0$.